The saucer for Roger's teacup has a radius of 3 centimeters. What is the saucer's circumference?

Mathematics

1 answer:

Fantom [35]3 years ago

7 0

C=dπ and r+r=d so 3+3=6 and 6π≈18.84 so the circumference would be 18.84

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2 years ago

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Evgesh-ka [11]

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2 years ago

3^x= 3*2^x solve this equation

kompoz [17]

In the equation

$3^x = 3\cdot 2^x$

divide both sides by $2^x$ to get

$\dfrac{3^x}{2^x} = 3 \cdot \dfrac{2^x}{2^x} \\\\ \implies \left(\dfrac32\right)^x = 3$

Take the base-3/2 logarithm of both sides:

$\log_{3/2}\left(\dfrac32\right)^x = \log_{3/2}(3) \\\\ \implies x \log_{3/2}\left(\dfrac 32\right) = \log_{3/2}(3) \\\\ \implies \boxed{x = \log_{3/2}(3)}$

Alternatively, you can divide both sides by $3^x$ :

$\dfrac{3^x}{3^x} = \dfrac{3\cdot 2^x}{3^x} \\\\ \implies 1 = 3 \cdot\left(\dfrac23\right)^x \\\\ \implies \left(\dfrac23\right)^x = \dfrac13$

Then take the base-2/3 logarith of both sides to get

$\log_{2/3}\left(2/3\right)^x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x \log_{2/3}\left(\dfrac23\right) = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(\dfrac13\right) \\\\ \implies x = \log_{2/3}\left(3^{-1}\right) \\\\ \implies \boxed{x = -\log_{2/3}(3)}$